Difference between revisions of "2010 AMC 8 Problems/Problem 12"
(→Solution 1) |
Prezlink16 (talk | contribs) (→Solution 3) |
||
| (9 intermediate revisions by 5 users not shown) | |||
| Line 7: | Line 7: | ||
Since 80 percent of the 500 balls are red, there are 400 red balls. Therefore, there must be 100 blue balls. For the 100 blue balls to be 25% or <math>\dfrac{1}{4}</math> of the bag, there must be 400 balls in the bag so 100 red balls must be removed. The answer is <math>\boxed{\textbf{(D)}\ 100}</math>. | Since 80 percent of the 500 balls are red, there are 400 red balls. Therefore, there must be 100 blue balls. For the 100 blue balls to be 25% or <math>\dfrac{1}{4}</math> of the bag, there must be 400 balls in the bag so 100 red balls must be removed. The answer is <math>\boxed{\textbf{(D)}\ 100}</math>. | ||
| − | ==Solution 2== | + | ==Solution 2(algebra solution)== |
We could also set up a proportion. Since we know there are 400 red balls, we let the amount of red balls removed be <math>x</math>, so <math>\frac{400-x}{500-x}=\frac{3}{4}</math>. Cross-multiplying gives us <math>1600-4x=1500-3x \implies x=100</math>, so our answer is <math>\boxed{\textbf{(D)}\ 100}</math>. | We could also set up a proportion. Since we know there are 400 red balls, we let the amount of red balls removed be <math>x</math>, so <math>\frac{400-x}{500-x}=\frac{3}{4}</math>. Cross-multiplying gives us <math>1600-4x=1500-3x \implies x=100</math>, so our answer is <math>\boxed{\textbf{(D)}\ 100}</math>. | ||
| + | |||
| + | ==Solution 3== | ||
| + | Before any balls are taken out of the bag, there are 400 red balls. Let the number of red balls we take out of the bag be <math>x</math>. Since we want the number of red ball to be 75% of the remaining balls, we can set up the equation: | ||
| + | <math>0.75(500–x)=400–x</math>. | ||
| + | In words, 75% of the total remaining balls are equal to the remaining red balls. | ||
| + | Solving we get <math>x=100</math>, so our answer is <math>\boxed{\textbf{(D)}\ 100}</math>. | ||
| + | ~J.L.L | ||
| + | |||
| + | ==Video by MathTalks== | ||
| + | |||
| + | https://www.youtube.com/watch?v=6hRHZxSieKc | ||
| + | |||
| + | |||
| + | |||
| + | |||
| + | |||
| + | |||
==See Also== | ==See Also== | ||
| + | |||
{{AMC8 box|year=2010|num-b=11|num-a=13}} | {{AMC8 box|year=2010|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 13:46, 19 August 2023
Contents
Problem
Of the 
balls in a large bag, 
are red and the rest are blue. How many of the red balls must be removed so that 
of the remaining balls are red?
Solution 1(logical solution)
Since 80 percent of the 500 balls are red, there are 400 red balls. Therefore, there must be 100 blue balls. For the 100 blue balls to be 25% or 
of the bag, there must be 400 balls in the bag so 100 red balls must be removed. The answer is 
.
Solution 2(algebra solution)
We could also set up a proportion. Since we know there are 400 red balls, we let the amount of red balls removed be 
, so 
. Cross-multiplying gives us 
, so our answer is .
Solution 3
Before any balls are taken out of the bag, there are 400 red balls. Let the number of red balls we take out of the bag be . Since we want the number of red ball to be 75% of the remaining balls, we can set up the equation:
.
In words, 75% of the total remaining balls are equal to the remaining red balls.
Solving we get 
, so our answer is .
~J.L.L
Video by MathTalks
https://www.youtube.com/watch?v=6hRHZxSieKc
See Also
| 2010 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
